What are the solutions to the equation a^2-2a+1=0 ?

The solutions to the equation a^2-2a+1=0 are a=1

Find the zeros of a^2-2a+1=0

The zeros of a^2-2a+1=0 are a=1

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a^{2} - 2 a + 1 = 0

a = 1

Solution steps

a^{2} - 2 a + 1 = 0

Solve with the quadratic formula

a_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 1 \cdot 1}{2 \cdot 1}

(- 2)^{2} - 4 \cdot 1 \cdot 1 = 0

a_{1, 2} = \frac{- ( - 2 ) \pm 0}{2 \cdot 1}

a = \frac{- ( - 2 )}{2 \cdot 1}

\frac{- ( - 2 )}{2 \cdot 1} = 1

a = 1

The solution to the quadratic equation is:

a = 1

347950 = x \cdot [\frac{1 - ( 1 + \frac{0.0045}{52} ) ^{- 260}}{\frac{0.0045}{52}}]

\frac{2 π}{3} - π

- x - 3 = 0

3 t^{2} - 12 t - 15 = 0

y^{4} + x^{2} y^{2}

What are the solutions to the equation a^2-2a+1=0 ?
The solutions to the equation a^2-2a+1=0 are a=1
Find the zeros of a^2-2a+1=0
The zeros of a^2-2a+1=0 are a=1